800 = 50 × e^(10k) → e^(10k) = 16 → 10k = ln(16) = 2.7726 → k ≈ 0.27726.
["Understanding the Mathematical Equation: 800 = 50 × e^(10k) → Step-by-Step Breakdown", "When dealing with exponential equations, transforming them into a more manageable form can reveal key insights and simplify calculations. One such equation is:", "[\n800 = 50 \ imes e^{10k}\n]", "This equation famously demonstrates how exponents and logarithms help solve for unknown variables. Let’s explore how to solve it step-by-step and understand its precise value.", "---", "### Step 1: Isolate the Exponential Term", "Start by dividing both sides of the equation by 50 to isolate the exponential expression:", "[\n\frac{800}{50} = e^{10k}\n]", "[\n16 = e^{10k}\n]", "This simplifies the original equation to a clean exponential form.", "---", "### Step 2: Apply Natural Logarithms to Both Sides", "To eliminate the exponent, we take the natural logarithm (ln) of both sides:", "[\n\ln(16) = \ln(e^{10k})\n]", "Using the logarithmic identity (\ln(e^x) = x), the right-hand side simplifies to:", "[\n\ln(16) = 10k\n]", "---", "### Step 3: Solve for (k)", "Now divide both sides by 10 to isolate (k):", "[\nk = \frac{\ln(16)}{10}\n]", "---", "### Step 4: Calculate Numerical Value", "We know from logarithmic tables or calculators that:", "[\n\ln(16) = \ln(2^4) = 4\ln(2) \approx 4 \ imes 0.693147 = 2.772588\n]", "Thus,", "[\nk = \frac{2.772588}{10} = 0.2772588\n]", "Rounded to five decimal places:", "[\nk \approx 0.27726\n]", "---", "### Summary of Key Values", "| Expression | Value |\n|-----------------------|---------------------|\n| 800 | 800 |\n| 50 | 50 |\n| Base of exponential | ( e ) |\n| Exponent | ( 10k ) |\n| (\ln(16)) | ≈ 2.7726 |\n| (10k) | ≈ 2.7726 |\n| (k) | ≈ 0.27726 |", "---", "### Why This Matters: Practical Applications", "This type of exponential relationship appears in many real-world scenarios, such as:", "- Compound Interest: ( A = P e^{rt} ) models growth over time.\n- Radioactive Decay: Decay rates follow similar exponential functions.\n- Population Growth: Models exponential population increases under ideal conditions.", "Understanding how to manipulate equations like ( a = b \ imes e^{kt} ) allows scientists, engineers, and financial analysts to compute time-dependent variables efficiently.", "---", "### Conclusion", "By systematically isolating the exponential term and applying logarithms, we turned a complicated exponential equation into a straightforward calculation. The precise value, ( k \approx 0.27726 ), unlocks deeper understanding and practical use in modeling real-life growth and decay phenomena. Whether in physics, finance, or biology, mastery of such equations is essential for accurate analysis and predictive modeling.", "If you found this explanation helpful, share it to spread clarity on exponential mathematics!", "---", "Keywords: exponential equation 800 = 50 × e^(10k), solve for k, natural logarithm, ln(16), k ≈ 0.27726, e^(10k) = 16, mathematical derivation, logarithmic calculation, real-world applications of exponential functions."]
